# Maths Assignment

**Topics:**Differential equation, Mathematics, Terminal velocity

**Pages:**9 (1521 words)

**Published:**August 30, 2013

Item 4B

Rachel Reiser

Maths C

Rachel Reiser

Maths C

Question 1

ab1+f'(x)2 dx

y = acosh(xa)

If:

coshx=12ex+e-x

Then:

cosh(xa) = 12(exa+e-xa)

y = acosh(xa)

∴

y=a(exa+e-xa)2

y=a(exa+e-xa)2

dydx=f'x=ddxa(exa+e-xa)2

dydx=f'x=ddx12aexa+e-xa

f'x=12a1aexa+-1ae-xa

f'x=exa-e-xa2

f'x2=exa-e-xa22

f'x2=(12exa-12e-xa)(12exa-12e-xa)

f'x2=14e2xa-14e0-14e0+14e-2xa

f'x2=14e2xa-12+14e-2xa

f'x2=14e2xa-2+e-2xa

Assuming the catenary is symmetrical, the entire length of the wire is equal to two multiplied by the length between x = b and x = a (or x = 0) Entire arc length =2×0b1+f'(x)2 dx

Substitute the value for f'x2 into the formula above.

L =2×0b1+14e2xa-2+e-2xa dx

L =2×0b44+14e2xa-2+e-2xa dx

L =2×0b144+e2xa-2+e-2xa dx

L =2×0b142+e2xa+e-2xa dx

L=2×0b142+e2xa+e-2xa dx

L=2×140b2+e2xa+e-2xa dx

L=2×120b2+e2xa+e-2xa dx

Note: rule of perfect squares 2+e2xa+e-2xa=(exa+e-xa)2

L=0b(exa+e-xa)2 dx

L=0bexa+e-xa dx

L=aexa-ae-xa0b

L=aeba-ae-ba-(ae0a-ae-0a)

L=aeba-ae-ba-(a-a)

L=aeba-ae-ba

L=a(eba-e-ba)

∴ The length of the arc in terms of a and b is L=aeba-e-ba

Question 2

mdvdt=mg-kv2

Where:

m=mass of body

v=velocity

g=acceleration due to gravity (9.8m/s)

k=constant related to size and shape of object

mdvdt=mg-kv2

mdv=mg-kv2 ×dt

mdvmg-kv2=dt

dt=mmg-kv2dv

dt=m1mg-kv2dv

mg-kv2 may also be expressed as mg2-kv22

=mg2-kv22

=mg2-kv2

∴ Difference of two squares:

mg+kvmg-kv

∴dt=m1mg+kvmg-kvdv

RHS=m1mg+kvmg-kvdv

let 1mg+kvmg-kv=Amg+vk+Bmg-vk

∴1+0v=mg-vkA+mg+vkB

1+0v=mgA-vkA+mgB+vkB

1+0v=mgA+mgB+vkB-kA

∴1=mgA+mgB0=kB-kA

1=mg(A+B)0=k(B-A)

A+B=1mg --- 1 0=B-A --- 2

A=B --- 2

Sub 2 into 1:

2A=1mg

A=12mg

∴A=12mg B=12mg

1mg+kvmg-kv=Amg+vk+Bmg-vk

∴1mg+kvmg-kv=12mgmg+vk+12mgmg-vk

1mg+kvmg-kv=12mgmg+vk+12mgmg-vk

1mg+kvmg-kvdv=12mgmg+vkdv+12mgmg-vkdv

dt=m1mg+kvmg-kvdv

∴dt=m12mgmg+vk+12mgmg-vkdv

dt=m12mg+2vmgk+12mg-2vmgkdv

dt=m21mg+vmgk+1mg-vmgkdv

t+c=m21mgklnmg+vmgk-1mgklnmg-vmgk

t+c=m21mgklnmg+vmgk-lnmg-vmgk

t+c=m21mgklnmg+vmgkmg-vmgk

t+c=m2mgklnmg+vmgkmg-vmgk

Find c by subbing in v=0, t=0

0+c=m2mgklnmg+0mgkmg-0mgk

c=m2mgklnmgmg

c=m2mgkln1

c=m2mgk×0

∴c=0

∴t=m2mgklnmg+vmgkmg-vmgk

2tmgkm=lnmg+vmgkmg-vmgk

2tmgkm2=lnmg+vmgkmg-vmgk

2tgkm=lnmg+vmgkmg-vmgk

e2tgkm=mg+vmgkmg-vmgk

(mg-vmgk)e2tgkm=mg+vmgk

mge2tgkm-vmgke2tgkm=mg+vmgk

vmgk+vmgke2tgkm=mge2tgkm-mg

vmgk+mgke2tgkm=mge2tgkm-1

v=mge2tgkm-1mgk+mgke2tgkm

v=mge2tgkm-1mgk1+e2tgkm

v=mg2e2tgkm-1mgk1+e2tgkm

v=mge2tgkm-1k1+e2tgkm

v=mgke2tgkm-1e2tgkm+1

v=limt→∞mgke2tgkm-1e2tgkm+1

v=limt→∞mgke2tgkm-1e2tgkm+1÷e2tgkme2tgkm

v=limt→∞mgk1-e-2tgkm1+e-2tgkm

v=limt→∞mgk1-e-∞1+e-∞

e-∞→0

v=mgk1-01+0

v=mgk×1

v=mgk

Therefore the limiting of the body is mgk.

Terminal velocity (i.e. acceleration = 0 m/s2)

Acceleration =dvdt

Where:

v=velocity

mdvdt=mg-kv2

When acceleration =o:

m×0=mg-kv2

0=mg-kv2

kv2=mg

v=mgk

* Let terminal velocity (v)= 12m/s2

Body mass (m) = 30kg

Gravity (g) = 9.8 m/s2

v=mgk

12=30×9.8k

122=294k

k=294144

k=2.04167

v=mgke2tgkm-1e2tgkm+1

Sub values for k, m and g into equation:

v=30×9.82.04167e2t9.8×2.0416730-1e2t9.8×2.0416730+1

Graphical analysis:

Effect of ‘k’ with respect to proportion principles

Terminal velocity:

v=mgk

Where the principle of proportions states that:

y∝x

∴y=kx

Where k is a constant.

v=mgk

In this case mg is the constant (i.e. has a known numerical value) Where y=v, x=k

y=kx

Translates to:

y=1xmg

Where y=y, k=mg, x=1k

∴as y∝1x

v=1k

Therefore terminal velocity is inversely proportional to k.

Question 3

t=0, P=12000

t=200, P=200

dPdt=110P1-P200001-αft

Where:

α=constant which takes into account the effect of introduced predators ft=function of time

General solution to differential equation given above for the case where f(t) = t: dP=110P1-P200001-αft dt

dP=P10-P22000001-αft dt

dPP10-P2200000=1-αft dt

Sub in f(t)...

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